quick sort的主要思路是partition, 就是选一个element作为pivot,比如总是选最右边的element,然后将array走一遍,使得最左边的subarray都是小于或者等于pivot的值,接着都是大于pivot的值,最后将第一个大于pivot的element和pivot互换。然后返回这个pivot的index,接着再recursively 去sort pivot左边的array和pivot右边的array。
Quick Sort使用了Divide and Concur的思想: 找一个基准数, 把小于基准数的数都放到基准数之前, 把大于基准数的数都放到基准数之后 Worst case: O(n^2) Average case: O(nlogN) 步骤: 初始的数组 Array a[]: 基准数: X = a[0] = 51 i 的值: i = 0 j 的值: j = 9 (a.length) Step 1: ...
QuickSort BinaryTree`s deep LeetCode~ListNode 快排 func QuickSort<T: Comparable>(dest:[T])->[T]{ guard dest.count > 1 else { return dest } let middle = dest[dest.count/2] let bigger = dest.filter { (t:T) -> Bool in return t > middle } let equal = dest.filter { (t:T) ...
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+ [排序](all/sort.go) 32 changes: 32 additions & 0 deletions 32 all/sort.go Original file line numberDiff line numberDiff line change @@ -0,0 +1,32 @@ package all import ( "math/rand" ) func QuickSort(a []int) { quickSort(a, 0, len(a)-1) } func quickSort(a []int...
quicksort就是选出某一个数在数组里面应该待的位置。上一篇已经提到了。所以这里就可以用quick sort来找第n-k的位置上的数。n为数组长度。 my solution: class Solution { public int findKthLargest(int[] nums, int k) { if(nums.length == 1) return nums[0]; ...
quicksort(head, nullptr);returnhead; } }; 两个指针,fast 指向大元素链表尾部,slow 指向小元素链表尾部,slow->next 是大元素链表头部。 遇到需要放在小元素链表的元素,我们就 slow = slow -> next,占用大元素链表头部的空间放置小元素。而原来的大元素链表头部的元素用值交换的放置放到链表末尾。
3. Use quickSort, 其实跟[LeetCode] 973. K Closest Points to Origin_Medium tag: Sort, heap, quickSort做法一样。 T: average O(n), worst case O(n ^ 2) classSolution:deftopKFrequent(self, nums: List[int], k: int) ->List[int]: ...
code generation (Chen et al., 2023a). On the other hand, large language models have been shown to be capable of describing their generated problem solutions in both text (Wei et al., 2022; Kojima et al., 2022; Zhou et al., 2023) and code (Gao et al., 2022; Chen et al., ...
参考quickSort use two pointers to decrease Time into O(n * lgn ) or O(n), 跟这个题目的做法一样[LeetCode] 973. K Closest Points to Origin_Medium tag: Sort, heap, quickSort code classSolution:deffindKthLargest(self, nums: List[int], k: int) ->int: ...